Transcript Chapter 13
Chapter 13
Multiple Integrals
by Zhian Liang
13.1 Double integrals over
rectangles
Recall the definition of definite integrals of
functions of a single variable
Suppose f(x) is defined on a interval [a,b].
Taking a partition P of [a, b] into subintervals:
a x x x
0
1
n 1
x b
n
Choose the points in [ x , x ] and let x i x i x i 1
i 1
i
Using the areas of the small rectangles to approximate the
areas of the curve sided echelons
and summing them, we have
n
(1)
f ( x ) x
*
i 1
i
i
P max{ x }
i
b
(2)
n
f ( x ) dx lim f ( x ) x
*
a
P 0
i 1
i
i
Double integral of a function of two variables defined
on a closed rectangle like the following
R [ a , b ] [ c , d ] {( x , y ) R | a x b , c y d }
2
Taking a partition of the rectangle
a x x x
0
1
m 1
c y y y
0
1
n 1
x b
m
y d
n
Choosing a point
*
*
ij
ij
( x , y ) in Rij and form the
double Riemann sum
(3)
m
n
i 1
j 1
f ( x , y )A
*
*
ij
ij
ij
(4) DEFINITION The double integral of f over the
rectangle R is defined as
m
n
i 1
j 1
f ( x , y ) dA lim f ( x , y ) A
P 0
if this limit exists
*
*
ij
ij
ij
Using Riemann sum can be approximately evaluate a double
integral as in the following example.
EXAMPLE 1 Find an approximate value for the integral
( x 3 y ) dA , where R {( x , y ) | 0 x 2 , 0 y 2},
2
R
by computing
the double Riemann sum with partition pines x=1 and x=3/2
*
*
ij
ij
and taking ( x , y ) to be the center of each rectangle.
Solution The partition is shown as above Figure. The area of
each subrectangle is A 12 , ( x , y ) is the center R ,
ij
*
*
ij
ij
ij
and f(x,y)=x-3y2. So the corresponding Riemann sum is
2
2
f ( x , y )A
i 1
j 1
*
*
ij
ij
ij
f ( x , y )A f ( x , y )A f ( x , y )A f ( x , y )A
*
*
11
11
11
*
*
12
12
12
*
*
21
21
21
f ( 12 , 54 ) A f ( 12 , 74 ) A f ( 32 , 54 ) A f ( 32 , 74 ) A
11
67
16
12
51 1
123 1
1
12 139
16 2
16 2
16 2
958 11 . 875
Thus we have
( x 3 y ) dA 11 . 875
2
R
21
22
*
*
22
22
22
Interpretation of double integrals as volumes
m
n
i 1
j 1
m
n
i 1
j 1
v f ( x , y )A
(5)
ij
*
*
ij
ij
ij
V f ( x , y )A
i 1
j 1
m
n
ij
ij
*
*
ij
(6) THEOREM If f ( x , y ) 0 and f is continuous on
the rectangle R, then the volume of the solid that
z f ( x, y )
lies above R and under the surface
is
V f ( x , y ) dA
R
R
EXAMPLE 2
Estimate the volume of the solid that lies above the square
[ 0 , 2 ] [ 0 , 2 ] and below the elliptic paraboloid z 16 x 2 y .
Use the partition of R into four squares and choose
( x , y ) to be the upper right corner of R .
Sketch the solid and the approximating rectangle boxes.
2
*
*
ij
ij
ij
2
Solution The partition and the graph of the function are
as the above. The area of each square is 1.Approximating
the volume by the Riemann sum, we have
V f (1,1) A f (1, 2 ) A f ( 2 ,1) A f ( 2 , 2 ) A
11
12
21
22
13 (1) 7 (1) 10 (1) 4 (1) 34
This is the volume of the approximating rectangular
boxes shown as above.
The properties of the double integrals
(7) [ f ( x , y ) g ( x , y )] dA
f ( x , y ) dA g ( x , y ) dA
R
R
(8) cf ( x , y ) dA c f ( x , y ) dA
R
R
(9) If
f ( x, y ) g ( x, y )
for all x , y R ,
then
f ( x , y ) dA g ( x , y ) dA
R
R
R
EXERCISES 13.1
Page 837
1. 3. 15.16
13.2 Iterated Integrals
The double integral can be obtained by evaluating two
single integrals.
The steps to calculate f ( x , y ) dA , where R [ a , b ] [ c , d ] :
R
Fix x to calculate
A ( x ) f ( x , y )dy
d
c
Then calculate
b
A ( x )dx
a
with respect to y .
(1) A ( x )dx [ f ( x , y )dy ]dx (called iterated integral)
(2)
b
b
a
a
d
c
f ( x , y )dy dx [ f ( x , y )dy ]dx
b
a
d
b
c
a
d
c
(3) Similarly
f ( x , y )dx dy [ f ( x , y )dx ]dy
d
c
b
a
d
c
b
a
EXAMPLE 1 Evaluate the iterated integrals
3 2
2
( a ) x y dy dx
0 1
(See the blackboard)
2
3
2
(b) x y dx dy
1
0
(4) Fubini’s Theorem If f is continuous on the
rectangle R {( x , y ) | a x b , c y d }, then
f ( x , y ) dA f ( x , y )dy dx f ( x , y )dx dy
b d
a c
d
c
b
a
R
More generally, this is true if we assume that f
is bounded on R , f is discontinuous only on
a finite number of smooth curves, and the iterated
integrals exist.
Interpret the double integral f ( x , y ) dA as the
volume V of the solid
R
where
V A ( x )dx
b
a
A ( x ) f ( x , y )dy
d
c
A ( x ) is the area of a cross-section of S in the
plane through x perpendicular to the x-axis.
Similarly
EXAMPLE 2
Evaluate the double integral
where
( x 3 y )dA
2
R
R {( x , y ) | 0 x 2 , 1 y 2}.
(See the blackboard)
EXAMPLE 3
Evaluate
y sin( xy ) dA
R
,where
R [1, 2 ] [ 0 , ]
Solution 1 If we first integrate with respect to x,
we get
y sin( xy ) dA y sin( xy )dxdy
2
0
R
1
x2
0
x 1
[ cos( xy )]
dy
( cos 2 y cos y ) dy
0
sin 2 y sin y 0 0
1
2
Solution 2 If we first integrate with respect to y, then
y sin( xy ) dA y sin( xy )dydx
2
R
1
2
1
1
x
0
[
1
x
[
1
x
2
1
2
1
[
2
2
2
1
1
yd (cos( xy )) dx
x
2
2
sin( x )
x
2
1
x
0
sin( x )
x
y cos( xy ) | cos( xy ) dy ]dx
cos( x )
sin( x )
1
0
cos( x )
x
dx
2
1
dx
1
2
x
0
sin( xy ) | ]dx
0
]dx 1
2
sin( x )
x
2
dx
2
1
cos( x )
x
dx
d sin( x )
x
sin( x )
x
2
1
2
1
sin( x )
x
2
dx
sin( x )
x
2
1
0
EXAMPLE 4 Find the volume of the solid S that is
bounded by the elliptic paraboloid x 2 y z 16 , the
plane y 2 and x 2 , and three coordinate planes.
2
2
Solution We first observe that S is the solid that lies
under the surface z 16 x 2 y and the above the
2
Square
R [ 0 , 2 ] [ 0 , 2 ].
2
Therefore,
V (16 x 2 y ) dA
2
2
R
(16 x 2 y ) dxdy
2
2
0
2
0
16 x
2
0
2
0
2
88
3
88
3
y
1
3
x 2y x
3
2
2
3
4 y dy
4
3
y
2
0
48
x2
x0
dy
If f ( x , y ) g ( x ) h ( y ) on R [ a , b ] [ c , d ] , then
g ( x ) h ( y ) dA g ( x )dx h ( y )dy
b
a
R
d
c
EXAMPLE 5 If R [ 0 , 2 ] [ 0 , 2 ] , then
sin x cos ydA sin xdx cos ydy
2
2
0
0
R
[ cos x ] [sin y ]
2
0
1 1 1
2
0
EXERCISES 13.2
Page 842
1(2), 6, 10, 16, 17,
13.3 Double integrals over general regions
To integrate over general regions like
R
which is bounded, being enclosed in a rectangular region R .
Then we define a new function F with domain R by
(1)
f ( x, y )
F ( x, y )
0
if
( x, y ) D
if
( x , y ) is in R but not in D
If F is integrable over R , then we say f is integrable
over D and we define the double integral of f over D by
(2)
where F
f ( x , y )dA F ( x , y )dA
D
R
is given by Equation 1.
Geometric interpretation
When
f ( x, y ) 0
R
The volume under f and above D equals to that under
F and above R.
Type I regions
D {( x , y ) | a x b , g ( x ) y g ( x )}
1
2
(3) If f is continuous on a type I region D such that
D {( x , y ) | a x b , g ( x ) y g ( x )}
1
then
f ( x , y )dA
b
a
D
g2 ( x )
g1 ( x )
2
f ( x , y )dydx
Type II regions
(4)
(5)
D {( x , y ) | c y d , h ( y ) x h ( y )}
1
D
f ( x , y )dA
d
c
h2 ( y )
2
f ( x , y )dxdy
h1 ( y )
where D is a type II region given by Equation 4
it is Type I region!
1
1
1 x
2x
2
2
( x 2 y ) dy dx
xy y
1
1
2
2
y 1 x
y2 x
2
dx
x (1 x ) (1 x ) 2 x 4 x dx
1
2
2
2
3
4
1
3 x x 2 x x 1 dx
1
4
3
2
1
1
3
2
x
5
4
3
2
1
x
32
15
5
x
4
x
3
x
Example 2 Find the volume of the solid that lies under the
paraboloid z x y and above the region D in the xy
2
2
-plane bounded by the line y 2 x and the parabola y x .
2
Type II
Type I
D {( x , y ) | 0 x 2 , x y 2 x }
2
D {( x , y ) | 0 y 4 ,
y
2
x
y}
Type I
Solution 1
( x y ) dA ( ( x y ) dy ) dx
2
D
2
0
y2 x
2
3
x
y
y
dx
3 y x2
1 6
14
4
1
( x x
3
2
0
3
3
0
x
3
2
5/2
1
x ) dx
2
3
15
y
y x
2
2
7
y
y
x y / 2
7/2
x
8
3
2
2
4
1
6
7
4
96
y
4
x x x
21
0
y/2
4
0
5
5
y
6
2
216
35
0
( x y ) dxdy
4
0
13
1
7
Type II4
dy
1
3
2
x
0
2
0
D
4
2x
( 2 x x x x ) dx
3
3
2
( x y ) dA
Solution 2
2
2
1
3
y
3/2
216
35
y
2
5/2
2
1
24
y
3
1
2
3
y dy
Example 3 Evaluate xydA , where D is the region bounded
by the line y x 1 and the parabola y 2 x 6 .
D
2
D as a type I
D as a type II
Solution We prefer to express D as a type II
D ( x, y ) | 2 y 4,
4
y 1
2
2
xydA
D
1
1
2
2
y
2
4
4
2
x y 1
4
dy
2
y ( y 1) (
1
1
2
24
36
x
y
2
xydxdy
2
5
4
y
y
6
5
4y
y
4
2
3
4
y
1
2
y
2
2y
3
3 x y 1
2
2
3
2
2
y
4
y
x
y
2
2
3
3 ) dy
2
2
8 y dy
4
2
Example 4 Find the volume of the tetrahedron bounded by
the planes x 2 y z 2 , x 2 y , x 0 , and z 0 .
x
x
D ( x , y ) | 0 x 1, y 1
2
2
D
D
Solution
1 1
( 2 x 2 y ) dydx
V ( 2 x 2 y )dA
2
0 x
2
D
2 y xy y
1
2
0
x
y 1
x
2
y
x
Here is wrong in the book!
dx
2
2 x x 1
1
0
x
2
1
x 2 x 1 dx
1
2
0
x x
3
3
2
1
x
0
1
3
x
2
2
x
2
2
x
x
2
4
dx
Example 5 Evaluate the iterated integral
1 1
2
sin( y )dydx
0
x
D ( x , y ) | 0 x 1, x y 1 ( x , y ) | 0 y 1, 0 x y
D as a type I
D as a type II
Solution If we try to evaluate the integral as it stands, we
are faced with the task of first evaluating sin( y )dy . But it is
2
impossible to do so in finite terms since sin( y )dy is not an
elementary function.(See the end of Section 7.6.) So we must
2
change the order of integration. This is accomplished by first
expressing the given iterated integral as a double integral.
sin( y )dydx sin( y ) dA
1 1
2
2
0 x
D
Where
D ( x , y ) | 0 x 1, x y 1
Using the alternative description of D,
we have
D ( x , y ) | 0 y 1, 0 x y
This enables us to evaluate the integral in the reverse
order:
sin( y )dydx sin( y ) dA
1 1
2
2
0 x
D
sin( y )dxdy
1
y
2
0 0
x sin( y ) dy
1
x y
2
x0
0
y sin( y ) dy
1
2
0
1
2
1
2
2
cos( y )
(1 cos 1)
1
0
Properties of double integrals
(6)
(7)
(8)
(9)
[ f ( x , y ) g ( x , y )] dA f ( x , y ) dA g ( x , y ) dA
D
D
D
cf ( x , y ) dA c f ( x , y ) dA
D
D
f ( x , y ) dA g ( x , y ) dA
D
D
if f ( x , y ) g ( x , y ) for
all ( x , y ) in D .
[ f ( x , y ) dA f ( x , y ) dA f ( x , y ) dA
D
if D D
D1
1
D where
2
D2
D and D
1
2
do not overlap except
perhaps on their boundaries like the following:
(10) 1dA
(11) If m
D
A( D )
, the area of region D.
f ( x , y ) M for all ( x , y ) D , then
mA ( D ) f ( x , y ) dA MA ( D )
D
Example 6
U se Property 11 to estimate the integral e
sin x cos y
D
where D is the disk with
center the region and radius 2.
1 sin x 1, and 1 cos x 1,
Solution Since
dA ,
we have
and therefore
1 sin x cos x 1,
e e
1
sin x cos x
e e
1
thus, using m=e-1=1/e, M=e, and A(D)=(2)2 in Property 11
we obtain
4
e
e
D
sin x cos y
dA 4 πe
Exercises 13.3
Page 850: 7, 9, 11, 33, 35
13.4 DBOUBLE INTEGRALS IN POLAR
COORDINATE
Suppose we want to evaluate a double integral f ( x , y ) dA ,
where R is one of regions shown in the following.
R
(a)
R {( r , ) | 0 r 1, 0 2 }
(b)R {( r , ) | 1 r 2 ,0
}
Recall from Section 9.4 that the polar coordinates
of a point related to the rectangular coordinates ( r , )
by the equations:
r x y
2
2
2
x r cos
y r sin
The regions in
the above
figure are
special cases
of a polar rectangle
R {( r , ) | a r b , }
Do the following partition (called polar partition)
The center of this subrectangle
R {( r , ) | r r r ,
i 1
ij
is
ri
*
1
1
( ri 1 ri )
2
j 1
i
*
j
(
j
j)
j 1
2
}
and the area is
A
1
2
r
1
2
(r r )
1
2
( r r )( r r )
ij
where
2
i
j
2
i
i
i 1
i 1
i 1
i
i 1
i
r r
r r r
r
2
j
2
*
i
1
2
i
j
i
j
i 1
j
The typical Riemann sum is
(1)
m
n
i 1
j 1
f ( r cos , r sin ) A
*
*
i
m
n
i 1
j 1
*
j
*
i
j
ij
f ( r cos , r sin )r r
*
i
*
j
*
i
*
j
*
i
i
j
If we write g ( r , ) rf ( r cos , r sin ) , then the
above Riemann sum can be written as
m
n
i 1
j 1
g ( r , ) r
*
i
*
j
i
j
which is the Riemann sum of the double integral
g ( r , )dr d
b
a
Therefore we have
m
n
i 1
j 1
m
n
i 1
j 1
f ( x , y ) dA lim f ( r cos , r sin ) A
R
P 0
*
*
i
j
*
i
lim g ( r , ) r
P 0
*
i
*
j
*
j
i
j
g ( r , )drd
b
a
f ( r cos , r sin )rdrd
b
a
ij
(2) Change to polar coordinates in a double
integral If f is continuous on a polar
rectangle R given by 0 a r b , ,
where 0 2 , then
f ( x , y ) dA f ( r cos , r sin )rdrd
R
b
a
Caution: Do not forget the factor r in (2)!
Example 1 Evaluate ( 3 x 4 y )dA , where R is the region
in the upper half-plane bounded by the circles
x y 1, and x y 4 .
2
R
2
Solution
2
2
2
The region R can be described as
R {( x , y ) | y 0 , 1 x y 4} {( r , ) | 1 r 2 , 0 }
2
2
( 3 x 4 y )dA 0 1 ( 3 r cos 4 r sin )rdrd
2
2
2
2
R
( 3 r cos 4 r sin )drd
0
2
2
3
2
1
r2
r cos r sin r 1 d 7 cos 15 sin d
15
7 cos
(1 cos 2 ) d
2
3
4
2
0
0
0
7 sin
15
15
sin 2
2
4
2
0
15
2
Example 2 Find the volume of the solid bounded
by the xy-plane and the paraboloid z 1 x y
2
2
D {( x , y ) | x y 1}
2
2
{( r , ) | 0 r 1, 0 2 }
V (1 x y )dA
2
2
D
2
(1 r )rdrd
0
1
2
0
2
d ( r r )dr
1
0
3
0
1
r
r
2
2
4 0
2
2
4
What we have done so far can be extended to the
complicated type of region shown in the following.
(3) If f is continuous on a polar
region of the form
D {( r , ) | , h ( ) r h ( )}
1
2
then
f ( x , y ) dA
D
h ( )
2
h1 ( )
f ( r cos , r sin )rdrd
Example 3 Use a double integral to find the area enclosed
by one loop of the four-leaved rose r cos 2
D {( r , ) |
4
, 0 r cos 2 }
4
/4
A ( D ) dA
/4
cos 2
0
rdrd
D
1
r
2
/4
1
2
1
cos 2
d
2
/4
/4
0
cos 2 d
/4
/4
2
/4
1 cos 4 d
4
1
1
sin 4
4
4
/4
/4
8
Example 4 Find the volume of the solid that lies under the
paraboloid z x y , above the xy plane, and inside
the cylinder x y 2 x .
2
2
2
2
Solution The solid lies above the disk, whose boundary circle
D {( r , ) |
2
2
, 0 r 2 cos }
/2
2 cos
V ( x y ) dA / 2 0
2
2
r rdrd
2
D
r
4
4
/2
8
/2
/2
0
2 cos
d
4 cos d
0
cos d
4
/2
1 cos 2
d
2
8
/2
2 [1 2 cos 2
0
/2
2
0
/2
4
1 cos 4 ] d
2
1
/2
3
1
2 sin 2 sin 4
8
2
0
3 3
2
2
2 2
Exercises 13.4
Page 856: 1, 4, 6, 7, 22